isomorphism

Definition

For a category $\mathcal{C}$, a morphism $f\in {\text{Hom}}_{\mathcal{C}}(X,Y)$ is called an isomorphism if there exists a morphism $g\in {\text{Hom}}_{\mathcal{C}}(Y,X)$ such that

$$g\circ f={\text{id}}_{X}f\circ g={\text{id}}_Y.$$

Properties of isomorphisms

Uniqueness of inverse

If $f$ is a morphism, then $g$ given the properties above is unique:

Proof

Let $f\in {\text{Hom}}_{\mathcal{C}}(X,Y)$ be an isomorphism and let $g,\tilde{g}\in {\text{Hom}}_{\mathcal{C}}(Y,X)$ be inverse functions such that

$$\begin{array}{rlr}g\circ f& ={\text{id}}_X& \tilde{g}\circ f={\text{id}}_X\\ f\circ g& ={\text{id}}_Y& f\circ \tilde{g}={\text{id}}_Y\end{array}$$

This means that $f\circ g=f\circ \tilde{g}$. Using (associative) composition again with $g$ gives

$$\begin{array}{rl}g\circ \left(f\circ g\right)& =g\circ \left(f\circ \tilde{g}\right)\\ \left(g\circ f\right)\circ g& =\left(g\circ f\right)\circ \tilde{g}\\ {\text{id}}_X\circ g& ={\text{id}}_X\circ \tilde{g}\\ g& =\tilde{g}\end{array}$$

Therefore, the inverse function is unique, and we can denote it by $f^{-1}$.

Inverse of inverse

If $f$ is an isomorphism, then so too is $f^{-1}$ and $(f^{-1}{)}^{-1}=f$.

Proof

Let $f\in {\text{Hom}}_{\mathcal{C}}(X,Y)$ be an isomorphism. Stated again, this means that $\exists f^{-1}\in {\text{Hom}}_{\mathcal{C}}(Y,X)$ such that

f^{-1} \circ f = \text{id}_X \quad \text{and} \quad f \circ f^{-1} = \text{id}_Y$$ To prove $f^{-1}$ is an isomorphism, we need to find a function $g \in \text{Hom}_\mathcal{C}(X,Y)$ such that

g \circ f^{-1} = \text{id}_Y \quad \text{and} \quad f^{-1} \circ g = \text{id}_X$$ Let $g=f$, thus $(f^{-1}{)}^{-1}=f$.

Composition of isomorphisms.

If $f\in {\text{Hom}}_{\mathcal{C}}(X,Y)$ and $g\in {\text{Hom}}_{\mathcal{C}}(Y,Z)$ are isomorphisms, then so too is $g\circ f\in {\text{Hom}}_{\mathcal{C}}(X,Z)$.

Proof

Let $f\in {\text{Hom}}_{\mathcal{C}}(X,Y)$ and $g\in {\text{Hom}}_{\mathcal{C}}(Y,Z)$ be isomorphisms. Then there exist morphisms $f^{-1}\in {\text{Hom}}_{\mathcal{C}}(Y,X)$ and $g^{-1}\in {\text{Hom}}_{\mathcal{C}}(Z,Y)$ such that

$$\begin{array}{rlr}f\circ f^{-1}& ={\text{id}}_Y& g\circ g^{-1}={\text{id}}_Z\\ f^{-1}\circ f& ={\text{id}}_X& g^{-1}\circ g={\text{id}}_Y\end{array}$$

Then for the morphism $g\circ f$, consider the morphism $f^{-1}\circ g^{-1}$.

$$\begin{array}{rl}(g\circ f)\circ (f^{-1}\circ g^{-1})& =g\circ (f\circ f^{-1})\circ g^{-1}\\ & =g\circ {\text{id}}_Y\circ g^{-1}\\ & =g\circ g^{-1}\\ & ={\text{id}}_Z\end{array}$$

and

$$\begin{array}{rl}(f^{-1}\circ g^{-1})\circ (g\circ f)& =f^{-1}(\circ g^{-1}\circ g)\circ f\\ & =f^{-1}\circ {\text{id}}_Y\circ f\\ & ={\text{id}}_X\end{array}$$

Therefore $(f^{-1}\circ g^{-1})=(g\circ f{)}^{-1}$ and $g\circ f$ is a morphism.